Trig Q help! From 1989 HSC past paper (1 Viewer)

LoveHateSchool · Apr 21, 2012

Not sure what question, but I've scanned where it appears on my worksheet. It's question 31 on work sheet, partly on first + second column, in the slightly different colour as its highlighted on my sheet.

Could not look it up as my past paper book does not go back so far, so assistance on this question is much appreciated and will be repped!

SpiralFlex · Apr 21, 2012

$GROWL$

$Perimeter ABCX,$

$Length AB + Arc BC+ Arc AC.$

$Length AB is already give, that's just r.$

$$Arc BC$\;= r\theta$

$Arc AC$\; = \frac{1}{2}*2*\pi*\frac{r}{2} \;$(Half the circumference of the circle)$

$Summing this up we get,$

$P=r+r\theta+r\frac{\pi}{2}$

[HR][/HR]

$Now the cross section is the area of 1 square unit. Clearly the area consists of a sector plus a semi circle.$

$$Area of the sector$\; = \frac{1}{2}r^2 \theta$

$$Area of semi circle$\; = \frac{1}{2} \pi (\frac{r}{2})^2=\frac{1}{8} \pi r^2$

$Summing these we will get 1,$

$\frac{1}{8} \pi r^2+\frac{1}{2}r^2 \theta=1$

$Times all sides by 8,$

$\pi r^2 +4r^2\theta=8$

$\theta=\frac{8-\pi r^2}{4r^2}$

$=\frac{2}{r^2}-\frac{\pi}{4}$

$Now we can substitute back into original equation of the perimeter.$

$P=r+r(\frac{2}{r^2}-\frac{\pi}{4})+r\frac{\pi}{2}$

$\therefore P=r+\frac{2}{r}-\frac{r \pi}{4}+r\frac{\pi}{2}$

$P=\frac{2}{r}+r(1+\frac{\pi}{4})$

[HR][/HR]

$$Least perimeter occurs when P'=0.$

$P'=-\frac{2}{r^2}+(1+\frac{\pi}{4})=0$

$\therefore (1+\frac{\pi}{4})=\frac{2}{r^2}$

$4+\pi=\frac{8}{r^2}$

$$Hence,$ \;r^2=\frac{8}{4+\pi}$

$r=1.06 \; $(Sorry don't have calculator have to do it in head)$$

$P is minimum when r = 1.06$

$P=\frac{2}{1.06}+(1.06)(1+\frac{\pi}{4})$

$\therefore P =\;$3.8-3.9ish?$$

$Substitute that back into,$

$$(Your answer of P)$\;=1.06+(1.06)\theta+\frac{(1.06) \pi}{2}$

$\theta=...$

Lol I typed the whole solutions then closed window TWICE. >_>

LoveHateSchool · Apr 21, 2012

Thank you Maffs robot!

Trig Q help! From 1989 HSC past paper (1 Viewer)

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